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Nurkadir007
@Nurkadir007
August 2022
1
7
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f(x) = √1-3x^2 +1/x^2+4. f(x)= (8-3x^6)^3 - x^2/5-x^2
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Answers & Comments
perova19721Tania9
1)f(x)=x³-x²-x+8
f`(x)=3x²-2x-1=0
D=4+12=16
x1=(2-4)/6=-1/3 x2=(2+4)/6=1
+ _ +
--------------------------------------
возр -1/3 убыв 1 возр
x∈(-∞;-1/3) U (1;∞)
2)f(x)=x³-6x²
f`(x)=3x²-12x=3x(x-4)=0
x=0 x=4
+ _ +
--------------------------------------
0 4
max min
ymax(0)=0 ymin(4)=64-96=-32
3)f(x)=1/3x³-4x
f`(x)=x²-4=(x-2)(x+2)=0
x=2∈[0;3] x=-2∉[0;3]
f(0)=0 max
f(2)=8/3-8=-16/3 min
f(3)=9-12=-3
4)f(x)=x³-3x
D(y)∈(-∞;∞)
f(-x)=-x³+3x=-(x³-3x) -нечетная
Точки пересечения с осями
0=0 у=0
х³-3х=0 х(х²-3)=0 х=0 х=-√3 х=√3
(0;0) (-√3;0) (√3;0)
f`(x)=3x²-3=3(x-1)(x+1)=0
x=-1 x=1
+ _ +
--------------------------------------
возр -1 убыв 1 возр
max min
ymax(-1)=2 ymin(1)=-2
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Answers & Comments
f`(x)=3x²-2x-1=0
D=4+12=16
x1=(2-4)/6=-1/3 x2=(2+4)/6=1
+ _ +
--------------------------------------
возр -1/3 убыв 1 возр
x∈(-∞;-1/3) U (1;∞)
2)f(x)=x³-6x²
f`(x)=3x²-12x=3x(x-4)=0
x=0 x=4
+ _ +
--------------------------------------
0 4
max min
ymax(0)=0 ymin(4)=64-96=-32
3)f(x)=1/3x³-4x
f`(x)=x²-4=(x-2)(x+2)=0
x=2∈[0;3] x=-2∉[0;3]
f(0)=0 max
f(2)=8/3-8=-16/3 min
f(3)=9-12=-3
4)f(x)=x³-3x
D(y)∈(-∞;∞)
f(-x)=-x³+3x=-(x³-3x) -нечетная
Точки пересечения с осями
0=0 у=0
х³-3х=0 х(х²-3)=0 х=0 х=-√3 х=√3
(0;0) (-√3;0) (√3;0)
f`(x)=3x²-3=3(x-1)(x+1)=0
x=-1 x=1
+ _ +
--------------------------------------
возр -1 убыв 1 возр
max min
ymax(-1)=2 ymin(1)=-2